First Problem Assignment
EECS 401
Due on January 12, 2007
PROBLEM 1 (15 points) Fully explain your answers to the following questions.
(a) If events A and B are mutually exclusive and collectively exhaustive, are Ac and Bc
mutually exclusive?
Solution Ac ∩ Bc = (A ∪ B)c = Ωc = ∅. Thus the events Ac and Bc are mutually
exclusive.
(b) If events A and B are mutually exclusive but not collectively exhaustive, are Ac and
Bc collectively exhaustive?
Solution Let C = (Ac ∪ Bc )c , that is the part that is not contained in Ac ∪ Bc . Using
De Morgan’s Law C = A ∩ B = ∅. Thus, there is nothing that is not a part of Ac or Bc .
Hence, Ac and Bc are collectively exhaustive.
(c) If events A and B are collectively exhaustive but not mutually exclusive, are Ac and
Bc collectively exhaustive?
Solution As in previous part, let C = (Ac ∪ Bc )c = A ∩ B which is not null. Thus, Ac
and Bc are not collectively exhaustive.
PROBLEM 2 (10 points) Joe is a fool with probability 0.6, a thief with probability 0.7, and
neither with probability 0.25.
(a) Determine the probability that he is a fool or a thief but not both.
Solution Let A be the event that Joe is a fool and B be the event that Joe is a thief. We
are given that
P(A) = 0.6
P(B) = 0.7
P((A ∪ B)c ) = 0.25
This implies:
1
Due on January 12, 2007
Solutions
First Problem Assignment
P A ∪ B = 1 − P (A ∪ B)c = 0.75
P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.55
The event that he is a fool or a thief but not both is given by (A ∩ Bc ) ∪ (Ac ∩ B). Looking
at the Venn diagram, the probability should be:
P (A ∩ Bc ) ∪ (Ac ∩ B) = P(A) + P(B) − 2P(A ∩ B) = 0.2
(1)
A
B
We can also derive this as follows
(A ∩ Bc ) ∪ (Ac ∩ B) = (A ∪ B) ∩ (A ∩ B)c
Thus,
P (A ∪ B) ∩ (A ∩ B)c = P(A ∪ B) + P (A ∩ B)c ) − P (A ∪ B) ∪ (A ∩ B)c
= P(A ∪ B) + 1 − P(A ∩ B) − P(S) = P(A ∪ B) − P(A ∩ B) = 0.2
This is the same expression as in (1).
(b) Determine the conditional probability that he is a thief, given that he is not a fool.
Solution We need to find P(B|Ac ). We know that
P(B|Ac ) =
=
0.7 − 0.55
=
1 − 0.6
P(B) − P(B ∩ A)
P(B ∩ Ac )
=
P(Ac )
1 − P(A)
0.375
2
Due on January 12, 2007
Solutions
First Problem Assignment
PROBLEM 3 (35 points) Express each of the following events in terms of the events A, B and
C as well as the operations of complementation, union and intersection:
(a) at least one of the events A, B , C occurs;
Solution
A∪B∪C
A
B
C
(b) at most one of the events A, B , C occurs;
Solution
[(A ∩ B) ∪ (B ∩ C) ∪ (C ∩ A)]c
A
B
C
(c) none of the events A, B , C occurs;
Solution
Ac ∩ Bc ∩ Cc
A
B
C
3
Due on January 12, 2007
Solutions
First Problem Assignment
(d) all three events A, B , C occur;
Solution
A∩B∩C
A
B
C
(e) exactly one of the events A, B , C occurs;
Solution
[A ∩ (Bc ∩ Cc )] ∪ [B ∩ (Cc ∩ Ac )] ∪ [C ∩ (Ac ∩ Bc )]
A
B
C
(f) events A and B occur, but not C ;
Solution
A ∩ B ∩ Cc
A
B
C
4
Due on January 12, 2007
Solutions
First Problem Assignment
(g) either event A occurs or, if not, then B also does not occur.
Solution
A ∪ (Ac ∩ Bc )
A
B
C
In each case draw the corresponding Venn diagram.
PROBLEM 4 (14 points) Suppose A and B are two events with known probabilities.
(a) Can you compute P(A ∪ B) in terms of P(A) and P(B)? If so, explain. If not, find
the largest and smallest possible values P(A ∪ B) can take in terms of P(A) and P(B)
and give examples in which these values can be attained (i.e., give upper and lower
bounds for P(A ∪ B).).
Solution No, P(A ∪ B) can not be computed in terms of P(A) and P(B) only. We know
that,
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Knowing only P(A) and P(B) does not give complete information about P(A ∩ B). However we know that
0 6 P(A ∩ B) and P(A ∩ B) 6 min P(A), P(B)
∴ max P(A), P(B) 6 P(A ∪ B) 6 P(A) + P(B)
The following cases can achieve the lower and the upper bound
A
B
A B
Upper bound on Union
Lower bound on Union
5
Due on January 12, 2007
Solutions
First Problem Assignment
(b) Repeat previous part with P(A ∩ B) instead of P(A ∪ B).
Solution As shown in the previous part
0 6 P(A ∩ B) 6 min P(A), P(B)
Also, note the case of part (a) that achieves the upper bound on P(A ∪ B) also achieves
the lower bound on P(A ∩ B). Similarly, the case of part (a) that achieves the lower bound
on P(A ∪ B) also achieves the upper bound on P(A ∩ B).
PROBLEM 5 (14 points) Prove that A ⊂ C if A ⊂ B and B ⊂ C . (In order to prove this you
need to argue that for every element x ∈ A it is true that x ∈ C and that there is at least one
element y ∈ C such that y 6∈ A.)
Solution
A ⊂ B means that ∀x ∈ A =⇒ x ∈ B and ∃y ∈ B s.t. y 6∈ A
(1)
B ⊂ C =⇒ ∀x ∈ B =⇒ x ∈ C and ∃y ∈ C s.t. y 6∈ B
(2)
From (1) and (2) we get that
x ∈ A means that x ∈ B and ∀x ∈ B, x ∈ C
Thus, ∀x ∈ A =⇒ x ∈ C
i.e.,
A⊆C
(3)
Furthermore from (2) we get that
∃y ∈ C s.t. y 6∈ B
(1) implies
y 6∈ B =⇒ y 6∈ A
i.e,
∃y ∈ C s.t. y 6∈ A
Combining this with (3) we get that A ⊂ C.
6
Due on January 12, 2007
Solutions
First Problem Assignment
PROBLEM 6 (12 points) Let A and B be two sets.
(a) Show that (Ac ∩ Bc )c = A ∪ B and (Ac ∪ Bc )c = A ∩ B
Solution
(Ac ∩ Bc )c = (Ac )c ∪ (Bc )c
=A∪B
by De Morgan’s Law
because (Ac )c = A
Similarly:
(Ac ∪ Bc )c = (Ac )c ∩ (Bc )c
=A∩B
by De Morgan’s Law
because (Ac )c = A
(b) Consider rolling a six-sided die once. Let A be the set of outcomes where an odd
number comes up. Let B be the set of outcomes where a 1 or a 2 comes up. Calculate
the sets on both sides of the equalities in part 1, and verify that the equalities hold.
Solution A = {1, 3, 5} and B = {1, 2}. Then Ac = {2, 4, 6} and Bc = {3, 4, 5, 6}.
Ac ∩ Bc = {4, 6}, thus (Ac ∩ Bc )c = {1, 2, 3, 5}.
A ∪ B = {1, 2, 3, 5}. Thus (Ac ∩ Bc )c = A ∪ B.
Similarly, we can show that (Ac ∪ Bc )c = A ∩ B = {1}.
7
Due on January 12, 2007